Let $G=SL(n,\mathbb R)$ with Lie algebra $\mathfrak{g}=\mathfrak{sl}(n,\mathbb R)$. The classical minimal parabolic subgroup $B$ consists of the upper triangular matrices.

The parabolic subgroups $P$ containing $B$ have $1-1$ correspondence to the subsets $\Pi_s$ of the set of simple roots $\Pi$. As an example, Consider $$ P=\left\{\begin{pmatrix}*&*&*&*&*&*\\*&*&*&*&*&*\\&&*&*&*&*\\&&*&*&*&*\\&&*&*&*&*\\ &&&&&* \end{pmatrix}\right\} $$ It is a parabolic subgroup corresponds to $\Pi_s=\{\epsilon_1-\epsilon_2,\epsilon_3-\epsilon_4,\epsilon_4-\epsilon_5\}$, where $\epsilon_j(\mathrm{diag}(a_1,\cdots a_n))=a_j$. The parabolic subalgebra of $P$ has Langlands decomposition $$ \mathfrak{q}=\mathfrak{m}\oplus\mathfrak{a}\oplus\mathfrak{n} $$ with $$ \mathfrak{a}=\left\{\mathrm{diag}(a,a,b,b,b,c), 2a+3b+c=0\right\}. $$ The Weyl group $$ W(G,A):=N_K(\mathfrak{a})/C_K(\mathfrak{a}), $$ where $K=SO(n)$ the associated maximal compact subgroup of $G$.

How to calculate $C_K(\mathfrak{a})$ and $N_K(\mathfrak{a})$? Is $W(G,A)=\{I\}$ in the above example?


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